I solve the problem with a number of Fibonacci (+ negative). Something like this:
fibonacci 0 = 0 fibonacci 1 = 1 fibonacci n | n == 0 = 1 | n > 0 = fibonacci (n - 1) + fibonacci (n - 2) | n < 0 = fibonacci (n + 2) - fibonacci (n + 1) need to increase the speed of rendering. for this, you naturally need a battery.
How to do it?
Most correctly, of course, use the ready-made formula of the n th member :) But for the case when you need to find it programmatically, the following idea fits (I am writing an algorithm, I do not know Haskell's syntax):
fib n = helper 0 1 n where helper curr prev n | n == 0 = curr | n > 0 = helper (curr+prev) curr (n-1) | n < 0 = helper prev (curr-prev) (n+1) It calculates the sequence in linear time with tail recursion.
fibhelper would be worth making fibhelper local identifier, not a global one. Something like this: fib n = helper 0 1 n where helper curr prev n = ... - Pavel MayorovSource: https://ru.stackoverflow.com/questions/455647/
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